If $x$, $y$, and $z$ are positive with $xy=24$, $xz = 48$, and $yz=72$, what is the value of $x+y+z$?
Answer: Since $$x=\frac{24}{y}=\frac{48}{z}$$ we have $z = 2y$. So $72
= 2y^2$, which implies that $y=6$, $x = 4$, and $z = 12$. Hence $x+y+z = \boxed{22}$.